Problem 60 - A Complete Resource Book in Chemistry for JEE Main

Question

The frequency of radiation emitted when the electron falls from \(\mathrm{n}=4\) to \(\mathrm{n}=1\) in a hydrogen atom will be (Given ionization energy of \(\mathrm{H}=2.18 \times 10^{18} \mathrm{~J}\) atom \(^{-1}\) and \(h=6.625 \times 10^{-34} \mathrm{Js}\) ) (a) \(1.54 \times 10^{15} \mathrm{~s}^{-1}\) (b) \(1.03 \times 10^{15} \mathrm{~s}^{-1}\) (c) \(3.08 \times 10^{15} \mathrm{~s}^{-1}\) (d) \(2.00 \times 10^{15} \mathrm{~s}^{-1}\)

Step-by-Step Solution

Answer

The frequency is \( 3.08 \times 10^{15} \mathrm{~s}^{-1} \) (option c).

Step 1: Understand the Problem

We need to find the frequency of radiation emitted when an electron in a hydrogen atom transitions from energy level \( n = 4 \) to \( n = 1 \). This requires using the ionization energy given and Planck's constant.

Step 2: Calculate Energy Levels

The energy of an electron in a hydrogen atom at a level \( n \) is given by \( E_n = -\dfrac{2.18 \times 10^{-18}}{n^2} \) J. Compute \( E_1 = -2.18 \times 10^{-18} \) J and \( E_4 = -\dfrac{2.18 \times 10^{-18}}{16} \) J.

Step 3: Calculate Energy Difference

Calculate the energy difference \( \Delta E \) for the transition from \( n = 4 \) to \( n = 1 \). \( \Delta E = E_1 - E_4 \). Substitute the values: \( \Delta E = -2.18 \times 10^{-18} - (-1.3625 \times 10^{-19}) = 2.04375 \times 10^{-18} \) J.

Step 4: Use Planck's Formula to Find Frequency

Use the formula \( \Delta E = h u \) to find the frequency \( u \). Substitute \( \Delta E = 2.04375 \times 10^{-18} \) J and \( h = 6.625 \times 10^{-34} \) Js into the formula, and solve for \( u \): \[ u = \dfrac{2.04375 \times 10^{-18}}{6.625 \times 10^{-34}} \].

Step 5: Calculate and Identify Answer

Calculate \( u \) using the values from Step 4: \[ u = \dfrac{2.04375 \times 10^{-18}}{6.625 \times 10^{-34}} = 3.08 \times 10^{15} \] \( \mathrm{s}^{-1} \). Therefore, the correct answer is (c) \( 3.08 \times 10^{15} \mathrm{~s}^{-1} \).

Key Concepts

Hydrogen Atom Energy Levels

In the Bohr model, the hydrogen atom's electron exists in distinct energy levels denoted as "n". These levels are crucial in determining the behavior of the electron within the atom. The principal quantum number "n" can take on positive integer values, such as 1, 2, 3, and so on. The lower the "n" value, the closer the electron is to the nucleus and the lower its energy. Therefore, when an electron moves from a higher energy level to a lower one, it releases energy in the form of light or radiation. This phenomenon is key in analyzing electron transitions.

The energy for an electron at a certain level is given by the formula:

• \(E_n = -\dfrac{2.18 \times 10^{-18}}{n^2}\)

This equation shows that the energy is negatively proportional to the square of the energy level. For example, at the first energy level \( (n = 1) \), the energy is more negative compared to higher levels, indicating a more tightly bound electron. Conversely, at higher levels like \( n = 4 \), the energy is less negative, reflecting the electron's relatively freer state.

Understanding these energy levels helps predict the wavelengths and frequencies of radiation emitted during electron transitions, as energy differences correlate directly with the electromagnetic spectrum.

Electron Transition Frequency

When electrons transition between energy levels in the hydrogen atom, they emit or absorb specific frequencies of radiation. The frequency of this radiation is determined by the energy difference between the two energy levels involved in the transition.

The energy difference \( (\Delta E) \) is given by the change in energy as the electron moves from initial level \( n_i \) to a final level \( n_f \). The formula used is:

• \( \Delta E = E_{n_f} - E_{n_i} \)

After computing \( \Delta E \), the frequency of the emitted or absorbed radiation can be determined using the relation:

• \( \Delta E = h \cdot u \)

where \( u \) is the frequency of the radiation and \( h \) is Planck's constant. By substituting \( \Delta E \) and \( h \) into this equation, we can solve for \( u \).

This process allows us to calculate the frequency of light emitted when an electron drops from a higher to a lower orbit. The frequency tells us how energetic the transition is, and it is directly related to the light's color or position in the electromagnetic spectrum. This concept is fundamental in spectroscopy, where scientists analyze these frequencies to ascertain the chemical composition of substances.

Planck's Constant

Planck's constant, denoted \( h \), is a fundamental constant in quantum mechanics. It is essential in understanding the quantization of energy, specifically in the study of photons and electromagnetic radiation. Planck's constant is valued at \( 6.625 \times 10^{-34} \) Js and plays a pivotal role in the equation \( \Delta E = h \cdot u \), connecting the frequency of radiation to the energy change during electron transitions.

This constant is named after Max Planck, who first introduced the concept in his work on blackbody radiation. He proposed that energy is not continuous but rather comes in discrete packets, or "quanta." This was a significant breakthrough in the field of physics, shifting the paradigm from classical to quantum mechanics.

Planck's constant allows us to determine the energy associated with a photon of light given its frequency. The relationship \( E = h \cdot u \) is incredibly powerful, as it underpins the entire theory of quantum mechanics and the behavior of atomic and subatomic particles.

• Quantum Mechanics: Describes the rules of physics on a tiny scale.
• Photon Energy: Can be calculated using Planck’s relation.
• Essential for Calculating: Frequencies and energies of electromagnetic waves.

Understanding Planck’s constant enables scientists to unlock the mysteries of the atomic world, offering insights into everything from fundamental particle behavior to the nature of light itself.